ECEN 667 Power System Stability Lecture 17: Transient Stability Solutions, Load Models Prof. Tom Overbye Dept. of Electrical and Computer Engineering Texas A&M University, overbye@tamu.edu 1
Announcements Read Chapter 7 Homework 5 is due today Homework 6 is assigned today, due on Nov 9 Final is as per TAMU schedule. That is, Friday Dec 8 from 3 to 5pm 2
Nonlinear Network Equations With constant impedance loads the network equations can usually be written with I independent of V, then they can be solved directly (as we've been doing) 1 V Y Ιx ( ) In general this is not the case, with constant power loads one common example Hence a nonlinear solution with Newton's method is used We'll generalize the dependence on the algebraic variables, replacing V by y since they may include other values beyond just the bus voltages 3
Nonlinear Network Equations Just like in the power flow, the complex equations are rewritten, here as a real current and a reactive current YV I(x,y) = 0 The values for bus i are n g ( x,y) G V B V I 0 Di ik Dk ik QK NDi k1 n g ( x,y) G V B V I 0 Qi ik Qk ik DK NQi k1 This is a rectangular formulation; we also could have written the equations in polar form For each bus we add two new variables and two new equations If an infinite bus is modeled then its variables and equations are omitted since its voltage is fixed 4
Nonlinear Network Equations The network variables and equations are then n In general there G1 kvdk B1 kvqk I ND1( x,y) 0 are no slack buses k1 n VD1 GikVQk BikVDK INQ1( x,y) 0 k 1 V Q1 n V D2 G2kVDk B2kVQK I ND2( x,y) 0 y g( x, y) k1 V Dn n V Qn Gnk VDk BnkVQK I NDn( x,y) 0 k1 n GnkVQk BnkVDK I NQn( x,y) 0 k1 5
Nonlinear Network Equation Newton Solution The network equations are solved using a similar procedure to that of the Netwon-Raphson power flow Set v 0; make an initial guess of y, y ( v) While gy ( ) Do ( v1) ( v) ( v) 1 ( v) y y J( y ) g( y ) v v1 End While ( v) 6
Network Equation Jacobian Matrix The most computationally intensive part of the algorithm is determining and factoring the Jacobian matrix, J(y) Jy ( ) gd1( x, y) gd1( x, y) gd1( x, y) VD1 VQ1 V Qn gq1( x, y) gq1( x, y) gq1( x, y) V V V D1 Q1 Qn gqn( x, y) gqn( x, y) gqn( x, y) VD1 VQ1 V Qn 7
Network Jacobian Matrix The Jacobian matrix can be stored and computed using a 2 by 2 block matrix structure The portion of the 2 by 2 entries just from the Y bus are gˆdi ( x, y) gˆdi ( x, y) VDj V Qj Gij Bij gˆqi ( x, y) gˆqi ( x, y) Bij Gij VDj VQj The "hat" was added to the g functions to indicate it is just the portion from the Y bus The major source of the current vector voltage sensitivity comes from non-constant impedance loads; also dc transmission lines 8
Example: Constant Current and Constant Power Load As an example, assume the load at bus k is represented with a ZIP model P P P V P V P 2 Load, k BaseLoad, k z, k k i, k k p, k Q Q Q V Q V Q 2 Load, k BaseLoad, k z, k k i, k k p, k The constant impedance portion is embedded in the Y bus Pˆ P P V P P V P Load, k BaseLoad, k i, k k p, k BL, i, k k BL, p, k Qˆ Q Q V Q Q V Q Load, k BaseLoad, k i, k k p, k BL, i, k k BL, p, k Usually solved in per unit on network MVA base The base load values are set from the power flow 9
Example: Constant Current and The current is then Constant Power Load P ˆ ˆLoad, k jq Load, k I Load, k I D, Load, k jiq, Load, k V k 2 2 2 2 PBL, i, k VDK VQK PBL, p, k j QBL, i, k VDK VQK QBL, p, k V Dk jv Multiply the numerator and denominator by V DK +jv QK to write as the real current and the reactive current Qk * 10
I I Example: Constant Current and Constant Power Load V P V Q V P V Q Dk BL, p, k QK BL, p, k Dk BL, i, k QK BL, i, k D, Load, k 2 2 V 2 2 DK VQK VDK VQK V P V Q V P V Q Qk BL, p, k DK BL, p, k Qk BL, i, k DK BL, i, k Q, Load, k 2 2 V 2 2 DK VQK VDK VQK The Jacobian entries are then found by differentiating with respect to V DK and V QK Only affect the 2 by 2 block diagonal values Usually constant current and constant power models are replaced by a constant impedance model if the voltage goes too low, like during a fault 11
Example: 7.4.ZIP Case Example 7.4 is modified so the loads are represented by a model with 30% constant power, 30% constant current and 40% constant impedance In PowerWorld load models can be entered in a number of different ways; a tedious but simple approach is to specify a model for each individual load Right click on the load symbol to display the Load Options dialog, select Stability, and select WSCC to enter a ZIP model, in which p1&q1 are the normalized about of constant impedance load, p2&q2 the amount of constant current load, and p3&q3 the amount of constant power load Case is Example_7_4_ZIP 12
Example 7.4.ZIP One-line Bus 2 Bus 7 Bus 8 Bus 9 Bus 3 163 MW 7 Mvar 1.025 pu 1.026 pu 1.016 pu 1.032 pu 1.025 pu 85 MW -11 Mvar Bus 5 0.996 pu 100 MW 35 Mvar Bus 6 1.013 pu 125 MW 50 Mvar Bus 4 1.026 pu 90 MW 30 Mvar Bus1 1.040 pu slack 72 MW 27 Mvar 13
Example 7.4.ZIP Bus 8 Load Values As an example the values for bus 8 are given (per unit, 100 MVA base) BaseLoad, 8 BaseLoad, 8 BaseLoad, 8 BaseLoad, 8 2 1. 00 P 0. 4 1. 016 0. 31. 016 0. 3 P 0. 983 2 0. 35 Q 0. 4 1. 016 0. 31. 016 0. 3 Q 0. 344 1 j0. 35 I D, Load, 8 jiq, Load, 8 0. 9887 j0. 332 1. 0158 j0. 0129 * 14
Example: 7.4.ZIP Case For this case the 2 by 2 block between buses 8 and 7 is 1.155 9.784 9.784 1.155 And between 8 and 9 is 1.617 13.698 13.698 1.617 These entries are easily checked with the Y bus The 2 by 2 block for the bus 8 diagonal is 2.876 23.352 23.632 3.745 The check here is left for the student 15
Additional Comments When coding Jacobian values, a good way to check that the entries are correct is to make sure that for a small perturbation about the solution the Newton's method has quadratic convergence When running the simulation the Jacobian is actually seldom rebuilt and refactored If the Jacobian is not too bad it will still converge To converge Newton's method needs a good initial guess, which is usually the last time step solution Convergence can be an issue following large system disturbances, such as a fault 16
Simultaneous Implicit The other major solution approach is the simultaneous implicit in which the algebraic and differential equations are solved simultaneously This method has the advantage of being numerically stable 17
Simultaneous Implicit Recalling the second lecture, we covered two common implicit integration approaches for solving x f ( x) Backward Euler x( t t) x( t) tf x( t t) For a linear system we have x( t t) I t A x( t) 1 Trapezoidal t x( t t) x( t) ( t) ( t t) 2 f x f x For a linear system we have 1 t x( t t) I t A I A ( t) 2 x We'll just consider trapezoidal, but for nonlinear cases 18
Nonlinear Trapezoidal We can use Newton's method to solve the trapezoidal t x( t t) x( t) f x( t t) f x( t) 0 2 We are solving for x(t+t); x(t) is known The Jacobian matrix is f1 f1 x1 x n t J x( t t) I 2 fn f1 x1 x n x f ( x) with Right now we are just considering the differential equations; we'll introduce the algebraic equations shortly The I comes from differentiating -x(t+t) 19
Nonlinear Trapezoidal using Newton's Method The full solution would be at each time step Set the initial guess for x(t+t) as x(t), and initialize the iteration counter k = 0 Determine the mismatch at each iteration k as t h x x x f x f x 2 ( ) ( ) ( ) ( t t) k ( t t) k ( t) ( t t) k ( t) Determine the Jacobian matrix Solve Iterate until done ( k 1) ( k ) ( k ) 1 ( k ) x ( t t) x ( t t) J ( x ( t t) h x ( t t) 20
Infinite Bus GENCLS Example Use the previous two bus system with gen 4 again modeled with a classical model with X d '=0.3, H=3 and D=0 GENCLS Bus 1 X=0.22 Bus 2 Infinite Bus slack 11.59 Deg 1.095 pu 0.00 Deg 1.000 pu In this example Xth = (0.22 + 0.3), with the internal voltage തE 1 = 1.281 23.95 giving E' 1 =1.281 and d 1 = 23.95 21
Infinite Bus GENCLS Implicit Solution Assume a solid three phase fault is applied at the bus 1 generator terminal, reducing P E1 to zero during the fault, and then the fault is self-cleared at time T clear, resulting in the post-fault system being identical to the pre-fault system During the fault-on time the equations reduce to dd1 1, pus dt d1, pu 1 dt 23 10 That is, with a solid fault on the terminal of the generator, during the fault P E1 = 0 22
Infinite Bus GENCLS Implicit Solution The initial conditions are x( 0) d ( 0) 0. 418 ( ) pu 0 0 Let t = 0.02 seconds During the fault the Jacobian is 0. 02 0 s 1 3. 77 J x( t t) 2 0 0 I 0 1 Set the initial guess for x(0.02) as x(0), and f x 0 0 0. 1667 23
Infinite Bus GENCLS Implicit Solution Then calculate the initial mismatch 0. 02 h x x x f x f x 2 ( 0. 02) ( 0) ( 0. 02) ( 0) ( 0) ( 0. 02) ( 0) ( 0) With x(0.02) (0) = x(0) this becomes ( 0) 0. 418 0. 418 0. 02 0 0 0 hx( 0. 02) 0 0 2 0. 167 0. 167 0. 00334 Then ( 1) 0. 418 1 3. 77 0 0. 4306 x( 0. 02) 0 0 1 0. 00334 0. 00334 1 24
Infinite Bus GENCLS Implicit Solution Repeating for the next iteration f x 0. 02 1 1 259 0. 1667 ( ). ( 1) 0. 4306 0. 418 0. 02 1. 259 0 hx( 0. 02) 0. 00334 0 2 0. 167 0. 167 00. 00. Hence we have converged with x( 0. 02) 0. 4306 0. 00334 25
Infinite Bus GENCLS Implicit Solution Iteration continues until t = T clear, assumed to be 0.1 seconds in this example 0. 7321 x( 0. 10) 0. 0167 At this point, when the fault is self-cleared, the equations change, requiring a re-evaluation of f(x(t clear )) dd pus dt d pu 1 1. 281 1 sind dt 6 0. 52 f x 6. 30 01. 0. 1078 26
Infinite Bus GENCLS Implicit Solution With the change in f(x) the Jacobian also changes ( 0). 0. 02 0 s 1 3 77 J x( 0. 12 ) 2 0. 305 0 I 0. 00305 1 Iteration for x(0.12) is as before, except using the new function and the new Jacobian 0. 02 h x x x f x f x 2 (. ) ( 0) (. ) ( 0) (. ) (. ) ( 0) 0 12 0 12 0 01 0 12 ( 0. 10 ) ( 1) 0. 7321 1 3. 77 0. 1257 0. 848 x( 0. 12) 0. 0167 0. 00305 1 0. 00216 0. 0142 1 This also converges quickly, with one or two iterations 27
Computational Considerations As presented for a large system most of the computation is associated with updating and factoring the Jacobian. But the Jacobian actually changes little and hence seldom needs to be rebuilt/factored Rather than using x(t) as the initial guess for x(t+t), prediction can be used when previous values are available x x x x ( ) ( t t) 0 ( t) ( t) ( t t) 28
Two Bus Results The below graph shows the generator angle for varying values of t; recall the implicit method is numerically stable 29
Adding the Algebraic Constraints Since the classical model can be formulated with all the values on the network reference frame, initially we just need to add the network equations We'll again formulate the network equations using the form Ι( x, y) Y V or Y V Ι( x, y) 0 As before the complex equations will be expressed using two real equations, with voltages and currents expressed in rectangular coordinates 30
Adding the Algebraic Constraints The network equations are as before n G1 kvdk B1 kvqk I ND1( x,y) 0 k1 n VD1 GikVQk BikVDK INQ1( x,y) 0 k 1 V Q1 n V D2 G2kVDk B2kVQK I ND2( x,y) 0 y g( x, y) k1 V Dn n V Qn Gnk VDk BnkVQK I NDn( x,y) 0 k1 n GnkVQk BnkVDK I NQn( x,y) 0 k1 31
Classical Model Coupling of x and y In the simultaneous implicit method x and y are determined simultaneously; hence in the Jacobian we need to determine the dependence of the network equations on x, and the state equations on y With the classical model the Norton current depends on x as Ei di 1 I, G jb Ni i i Rs, i jx d, i Rs, i jx d, i E je E cosd j cosd sind sind I I ji E j G jb Ni DNi QNi i i i i i Di Qi i i i I E G E B DNi Di i Qi i I E B E G QNi Di i Qi i Recall with the classical model E i is constant 32
Classical Model Coupling of x and y In the state equations the coupling with y is recognized by noting P P P E I E I Ei Di Di Qi Qi I ji E V j E V G jb Di Qi Di Di Qi Qi i i I E V G E V B Di Di Di i Qi Qi i I E V B E V G Qi Di Di i Qi Qi i E E V G E V B E E V B E V G Ei Di Di Di i Qi Qi i Qi Di Di i Qi Qi i 2 2 E E V G E E V G E V E V B Ei Di Di Di i Qi Qi Qi i Di Qi Qi Di i 33
Variables and Mismatch Equations In solving the Newton algorithm the variables now include x and y (recalling that here y is just the vector of the real and imaginary bus voltages The mismatch equations now include the state integration equations ( k ) h x( t t) t x x f x y f x y 2 And the algebraic equations ( k ) ( k ) ( k ) ( t t) ( t) ( t t), ( t t) ( t), ( t) g x( t t), y( t t) ( k ) ( k) 34
Jacobian Matrix Since the h(x,y) and g(x,y) are coupled, the Jacobian is J x( t t), y( t t) ( k) ( k) ( k ) ( k ) ( k ) ( k ) ( t t), ( t t) ( t t), ( t t) h x y h x y x y g x y g x y x y ( k ) ( k ) ( k ) ( k ) ( t t), ( t t) ( t t), ( t t) With the classical model the coupling is the Norton current at bus i depends on d i (i.e., x) and the electrical power (P Ei ) in the swing equation depends on V Di and V Qi (i.e., y) 35
Jacobian Matrix Entries The dependence of the Norton current injections on d is I EcosdG Esind B DNi i i i i i i I Ecosd B EsindG QNi i i i i i i I d I DNi i QNi d i EsindG Ecosd B i i i i i i Esind B EcosdG i i i i i i In the Jacobian the sign is flipped because we defined g x, y = Y V Ι(x, yሻ 36
Jacobian Matrix Entries The dependence of the swing equation on the generator terminal voltage is d P i i. pu s 1 P P D i, pu Mi Ei i i, pu 2Hi 2 2 Ei Di Di Di i Qi Qi Qi i Di Qi Qi Di i V V E E V G E E V G E V E V B i, pu Di i, pu Qi 1 2H 1 2H i i E G E B Di i Qi i E G E B Qi i Di i 37
Two Bus, Two Gen GENCLS Example We'll reconsider the two bus, two generator case from Lecture 16; fault at Bus 1, cleared after 0.06 seconds Initial conditions and Y bus are as covered in Lecture 16 GENCLS Bus 1 X=0.22 Bus 2 GENCLS slack 11.59 Deg 1.095 pu 0.00 Deg 1.000 pu PowerWorld Case B2_CLS_2Gen 38
Two Bus, Two Gen GENCLS Example Initial terminal voltages are V jv 1. 0726 j0. 22, V jv 1. 0 D1 Q1 D2 Q2 E 1. 28123. 95, E 0. 955 12. 08 1 2 1. 1709 j0. 52 I N1 1. 733 j3. 903 j0. 3 0. 9343 j0. 2 I N2 1 j4. 6714 j0. 2 1 0 j0. 333 j7. 879 j4. 545 Y YN 1 j4. 545 j9. 545 0 j0. 2 39
Two Bus, Two Gen Initial Jacobian d1 1 d2 1 VD1 VQ1 VD2 VQ2 d1 1 3. 77 0 0 0 0 0 0 1 0. 0076 1 0 0 0. 0029 0. 0065 0 0 d2 0 0 1 3. 77 0 0 0 0 2 0 0 0. 0039 1 0 0 0. 0008 0. 0039 I D1 3. 90 0 0 0 0 7. 879 0 4. 545 IQ1 1. 73 0 0 0 7. 879 0 4. 545 0 I D2 0 0 4. 67 0 0 4. 545 0 9. 545 IQ2 0 0 1. 00 0 4. 545 0 9. 545 0 40
Results Comparison The below graph compares the angle for the generator at bus 1 using t=0.02 between RK2 and the Implicit Trapezoidal; also Implicit with t=0.06 41
Four Bus Comparison GENCLS Bus 4 X=0.1 Bus 1 Bus 2 GENCLS slack 7.72 Deg 1.0551 pu X=0.1 Bus 3 X=0.2 2.31 Deg 1.005 pu 100 MW 50 Mvar -2.40 Deg 0.966 pu 0.00 Deg 1.000 pu 42
Four Bus Comparison Fault at Bus 3 for 0.12 seconds; self-cleared 43
Done with Transient Stability Solutions: On to Load Modeling Load modeling is certainly challenging! For large system models an aggregate load can consist of many thousands of individual devices The load is constantly changing, with key diurnal and temperature variations For example, a higher percentage of lighting load at night, more air conditioner load on hot days Load model behavior can be quite complex during the low voltages that may occur in transient stability Testing aggregate load models for extreme conditions is not feasible we need to wait for disturbances! 44
Load Modeling Traditionally load models have been divided into two groups Static: load is a algebraic function of bus voltage and sometimes frequency Dynamic: load is represented with a dynamic model, with induction motor models the most common The simplest load model is a static constant impedance Has been widely used Allowed the Y bus to be reduced, eliminating essentially all non-generator buses Presents no issues as voltage falls to zero Is rapidly falling out of favor 45
Load Modeling References Many papers and reports are available! A classic reference on load modeling is by the IEEE Task Force on Load Representation for Dynamic Performance, "Load Representation for Dynamic Performance Analysis," IEEE Trans. on Power Systems, May 1993, pp. 472-482 A more recent report that provides a good overview is "Final Project Report Loading Modeling Transmission Research" from Lawrence Berkeley National Lab, March 2010 46
ZIP Load Model Another common static load model is the ZIP, in which the load is represented as P P P V P V P Load, k BaseLoad, k z, k k i, k k p, k Q Q Q V Q V Q Load, k BaseLoad, k z, k k i, k k p, k 2 2 Some models allow more general voltage dependence n1 n2 n3 P P a V a V a V Load, k BaseLoad, k 1, k k 2, k k 3, k k n4 n5 n6 Q Q a V a V a V Load, k BaseLoad, k 4, k k 5, k k 6, k k The voltage exponent for reactive power is often > 2 47
ZIP Model Coefficients An interesting paper on the experimental determination of the ZIP parameters is A. Bokhari, et. al., "Experimental Determination of the ZIP Coefficients for Modern Residential and Commercial Loads, and Industrial Loads," IEEE Trans. Power Delivery, 2014 Presents test results for loads as voltage is varied; also highlights that load behavior changes with newer technologies Below figure (part of fig 4 of paper), compares real and reactive behavior of light ballast 48
ZIP Model Coefficients The Z,I,P coefficients sum to zero; note that for some models the absolute values of the parameters are quite large, indicating a difficult fit A portion of Table VII from Bokhari 2014 paper 49
Discharge Lighting Models Discharge lighting (such as fluorescent lamps) is a major portion of the load (10-15%) Discharge lighting has been modeled for sufficiently high voltage with a real power as constant current and reactive power with a high voltage dependence Linear reduction for voltage between 0.65 and 0.75 pu Extinguished (i.e., no load) for voltages below P P V Discharg elighting Base k Q Q V Discharg elighting Base k 45. May need to change with newer electronic ballasts e.g., reactive power increasing as the voltage drops! 50
Static Load Model Frequency Dependence Frequency dependence is sometimes included, to recognize that the load could change with the frequency 2 P P P V P V P 1 P f 1 Load, k BaseLoad, k z, k k i, k k p, k f, k k 2 Q Q Q V Q V Q 1 Q f 1 Load, k BaseLoad, k z, k k i, k k p, k f, k k Here f k is the per unit bus frequency, which is calculated as k s 1 st f k A typical value for T is about 0.02 seconds. Some models just have frequency dependence on the constant power load Typical values for P f and Q f are 1 and -1 respectively 51
Aside: Voltage Stability Next few slides are an aside on static voltage stability Voltage Stability: The ability to maintain system voltage so that both power and voltage are controllable. System voltage responds as expected (i.e., an increase in load causes proportional decrease in voltage). Voltage Instability: Inability to maintain system voltage. System voltage and/or power become uncontrollable. System voltage does not respond as expected. Voltage Collapse: Process by which voltage instability leads to unacceptably low voltages in a significant portion of the system. Typically results in loss of system load. 52
Voltage Stability Two good references are P. Kundur, et. al., Definitions and Classification of Power System Stability, IEEE Trans. on Power Systems, pp. 1387-1401, August 2004. T. Van Cutsem, Voltage Instability: Phenomena, Countermeasures, and Analysis Methods, Proc. IEEE, February 2000, pp. 208-227. Classified by either size of disturbance or duration Small or large disturbance: small disturbance is just perturbations about an equilibrium point (power flow) Short-term (several seconds) or long-term (many seconds to minutes) 53